Aspirin can be prepared by the reaction of acetyl chloride with :

Options

A. Benzoic acid
B. Phenol
C. p-hydroxy benzoic acid
D. o-hydroxy benzoic acid

Answer

To prepare Aspirin (acetylsalicylic acid), we need to react acetyl chloride with the appropriate starting material. The correct starting material for this reaction is ortho-hydroxybenzoic acid, also known as salicylic acid. Here’s a step-by-step breakdown of the reaction:

### Step-by-Step Solution:

1. **Identify the Reactants:**
- Acetyl chloride (CH3COCl) is the acetylating agent.
- The appropriate acid to react with is ortho-hydroxybenzoic acid (salicylic acid).

2. **Structure of Salicylic Acid:**
- Salicylic acid has the structure:
\[ \text{C}_6\text{H}_4(\text{OH})(\text{COOH}) \]
- The hydroxyl (–OH) group is in the ortho position relative to the carboxylic acid (–COOH) group.

3. **Reaction Mechanism:**
- When salicylic acid reacts with acetyl chloride, the hydroxyl group (–OH) of salicylic acid is protonated, making it a better leaving group.
- The acetyl chloride donates an acetyl group (–COCH3) to the salicylic acid.

4. **Formation of Aspirin:**
- The reaction proceeds as follows:
- The acetyl chloride reacts with the –OH group of salicylic acid, leading to the formation of aspirin (acetylsalicylic acid):
\[ \text{C}_6\text{H}_4(\text{OH})(\text{COOH}) + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_4(\text{OCOCH}_3)(\text{COOH}) + \text{HCl} \]
- The product formed is aspirin, which has the structure:
\[ \text{C}_6\text{H}_4(\text{OCOCH}_3)(\text{COOH}) \]

5. **Conclusion:**
- The correct answer is that aspirin can be prepared by the reaction of acetyl chloride with ortho-hydroxybenzoic acid (salicylic acid).

### Final Answer:
Aspirin can be prepared by the reaction of acetyl chloride with **ortho-hydroxybenzoic acid** (Option 4).

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